Is $C^1([a,b], mathbb{R}^n)$ a reflexive Banach space?












1












$begingroup$


I want to prove or disprove that $C^1([a,b], mathbb{R}^n)$ equipped with the norm $||x||=underset{tin[a,b]}{sup}|x(t)|_{mathbb{R}^n}+underset{tin[a,b]}{sup}|dot{x}(t)|_{mathbb{R}^n}$ is a reflexive Banach space.



I figured out that it is indeed a Banach space, but I'm stuck on the reflexive part. I know that a Banach space $X$ is reflexive if the mapping $F: X rightarrow X''$ is surjective, where $X''$ is the second dual space of $X$.



However, I have the feeling that this space is not reflexive, but I'm not sure how to prove that.



Is it indeed true that this space is not reflexive?






banach-spaces normed-spaces reflexive-space







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  • $begingroup$
    A proof is outlined here, although you'll have to put some pieces together: math.stackexchange.com/questions/891862/…
    $endgroup$
    – Olivier Moschetta
    Nov 27 '18 at 19:11
















1












$begingroup$


I want to prove or disprove that $C^1([a,b], mathbb{R}^n)$ equipped with the norm $||x||=underset{tin[a,b]}{sup}|x(t)|_{mathbb{R}^n}+underset{tin[a,b]}{sup}|dot{x}(t)|_{mathbb{R}^n}$ is a reflexive Banach space.



I figured out that it is indeed a Banach space, but I'm stuck on the reflexive part. I know that a Banach space $X$ is reflexive if the mapping $F: X rightarrow X''$ is surjective, where $X''$ is the second dual space of $X$.



However, I have the feeling that this space is not reflexive, but I'm not sure how to prove that.



Is it indeed true that this space is not reflexive?






banach-spaces normed-spaces reflexive-space







share|cite|improve this question











$endgroup$












  • $begingroup$
    A proof is outlined here, although you'll have to put some pieces together: math.stackexchange.com/questions/891862/…
    $endgroup$
    – Olivier Moschetta
    Nov 27 '18 at 19:11














1












1








1





$begingroup$


I want to prove or disprove that $C^1([a,b], mathbb{R}^n)$ equipped with the norm $||x||=underset{tin[a,b]}{sup}|x(t)|_{mathbb{R}^n}+underset{tin[a,b]}{sup}|dot{x}(t)|_{mathbb{R}^n}$ is a reflexive Banach space.



I figured out that it is indeed a Banach space, but I'm stuck on the reflexive part. I know that a Banach space $X$ is reflexive if the mapping $F: X rightarrow X''$ is surjective, where $X''$ is the second dual space of $X$.



However, I have the feeling that this space is not reflexive, but I'm not sure how to prove that.



Is it indeed true that this space is not reflexive?






banach-spaces normed-spaces reflexive-space







share|cite|improve this question











$endgroup$




I want to prove or disprove that $C^1([a,b], mathbb{R}^n)$ equipped with the norm $||x||=underset{tin[a,b]}{sup}|x(t)|_{mathbb{R}^n}+underset{tin[a,b]}{sup}|dot{x}(t)|_{mathbb{R}^n}$ is a reflexive Banach space.



I figured out that it is indeed a Banach space, but I'm stuck on the reflexive part. I know that a Banach space $X$ is reflexive if the mapping $F: X rightarrow X''$ is surjective, where $X''$ is the second dual space of $X$.



However, I have the feeling that this space is not reflexive, but I'm not sure how to prove that.



Is it indeed true that this space is not reflexive?






banach-spaces normed-spaces reflexive-space




banach-spaces normed-spaces reflexive-space






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 19:11









Davide Giraudo

126k16150261




126k16150261










asked Nov 27 '18 at 19:05









CatemathicsCatemathics

1013




1013












  • $begingroup$
    A proof is outlined here, although you'll have to put some pieces together: math.stackexchange.com/questions/891862/…
    $endgroup$
    – Olivier Moschetta
    Nov 27 '18 at 19:11


















  • $begingroup$
    A proof is outlined here, although you'll have to put some pieces together: math.stackexchange.com/questions/891862/…
    $endgroup$
    – Olivier Moschetta
    Nov 27 '18 at 19:11
















$begingroup$
A proof is outlined here, although you'll have to put some pieces together: math.stackexchange.com/questions/891862/…
$endgroup$
– Olivier Moschetta
Nov 27 '18 at 19:11




$begingroup$
A proof is outlined here, although you'll have to put some pieces together: math.stackexchange.com/questions/891862/…
$endgroup$
– Olivier Moschetta
Nov 27 '18 at 19:11










2 Answers
2






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oldest

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2












$begingroup$

There's a standard theorem that for any Banach space $X$, if $X'$ is separable then so is $X$. Hence, if $X''$ is separable then so is $X'$.



It is not so hard to show that if $X = C^1([a,b])$, then $X'$ is not separable. (I'll take $n=1$.) For instance, for any $t in [a,b]$, consider the linear functional $ell_t(x) = dot{x}(t)$. You can check that it is continuous, and that $|ell_t - ell_s|_{X'} = 2$ for any $s ne t$. Hence there is an uncountable closed discrete set in $X'$.



You can also check that $X$ itself is separable (using some version of Stone-Weierstrass). So if $X$ were reflexive, then $X'' = X$ would be separable, hence $X'$ would also be separable, but it isn't.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    We can embed $Y = C([a,b],mathbb R)$ into $X = C^1([a,b], mathbb R^n)$ by
    $T(f)(x) = int_a^x f(t); dt; {bf u}$ where $bf u$ is some nonzero vector in $mathbb R^n$. Then $T^{**}$ embeds $Y^{**}$ into $X^{**}$. But $Y^{**}$ is not separable, therefore $X^{**}$ is not separable, but $X$ is separable, so $X$ is not reflexive.






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

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      2












      $begingroup$

      There's a standard theorem that for any Banach space $X$, if $X'$ is separable then so is $X$. Hence, if $X''$ is separable then so is $X'$.



      It is not so hard to show that if $X = C^1([a,b])$, then $X'$ is not separable. (I'll take $n=1$.) For instance, for any $t in [a,b]$, consider the linear functional $ell_t(x) = dot{x}(t)$. You can check that it is continuous, and that $|ell_t - ell_s|_{X'} = 2$ for any $s ne t$. Hence there is an uncountable closed discrete set in $X'$.



      You can also check that $X$ itself is separable (using some version of Stone-Weierstrass). So if $X$ were reflexive, then $X'' = X$ would be separable, hence $X'$ would also be separable, but it isn't.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        There's a standard theorem that for any Banach space $X$, if $X'$ is separable then so is $X$. Hence, if $X''$ is separable then so is $X'$.



        It is not so hard to show that if $X = C^1([a,b])$, then $X'$ is not separable. (I'll take $n=1$.) For instance, for any $t in [a,b]$, consider the linear functional $ell_t(x) = dot{x}(t)$. You can check that it is continuous, and that $|ell_t - ell_s|_{X'} = 2$ for any $s ne t$. Hence there is an uncountable closed discrete set in $X'$.



        You can also check that $X$ itself is separable (using some version of Stone-Weierstrass). So if $X$ were reflexive, then $X'' = X$ would be separable, hence $X'$ would also be separable, but it isn't.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          There's a standard theorem that for any Banach space $X$, if $X'$ is separable then so is $X$. Hence, if $X''$ is separable then so is $X'$.



          It is not so hard to show that if $X = C^1([a,b])$, then $X'$ is not separable. (I'll take $n=1$.) For instance, for any $t in [a,b]$, consider the linear functional $ell_t(x) = dot{x}(t)$. You can check that it is continuous, and that $|ell_t - ell_s|_{X'} = 2$ for any $s ne t$. Hence there is an uncountable closed discrete set in $X'$.



          You can also check that $X$ itself is separable (using some version of Stone-Weierstrass). So if $X$ were reflexive, then $X'' = X$ would be separable, hence $X'$ would also be separable, but it isn't.






          share|cite|improve this answer









          $endgroup$



          There's a standard theorem that for any Banach space $X$, if $X'$ is separable then so is $X$. Hence, if $X''$ is separable then so is $X'$.



          It is not so hard to show that if $X = C^1([a,b])$, then $X'$ is not separable. (I'll take $n=1$.) For instance, for any $t in [a,b]$, consider the linear functional $ell_t(x) = dot{x}(t)$. You can check that it is continuous, and that $|ell_t - ell_s|_{X'} = 2$ for any $s ne t$. Hence there is an uncountable closed discrete set in $X'$.



          You can also check that $X$ itself is separable (using some version of Stone-Weierstrass). So if $X$ were reflexive, then $X'' = X$ would be separable, hence $X'$ would also be separable, but it isn't.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 '18 at 19:15









          Nate EldredgeNate Eldredge

          63k682171




          63k682171























              1












              $begingroup$

              We can embed $Y = C([a,b],mathbb R)$ into $X = C^1([a,b], mathbb R^n)$ by
              $T(f)(x) = int_a^x f(t); dt; {bf u}$ where $bf u$ is some nonzero vector in $mathbb R^n$. Then $T^{**}$ embeds $Y^{**}$ into $X^{**}$. But $Y^{**}$ is not separable, therefore $X^{**}$ is not separable, but $X$ is separable, so $X$ is not reflexive.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                We can embed $Y = C([a,b],mathbb R)$ into $X = C^1([a,b], mathbb R^n)$ by
                $T(f)(x) = int_a^x f(t); dt; {bf u}$ where $bf u$ is some nonzero vector in $mathbb R^n$. Then $T^{**}$ embeds $Y^{**}$ into $X^{**}$. But $Y^{**}$ is not separable, therefore $X^{**}$ is not separable, but $X$ is separable, so $X$ is not reflexive.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  We can embed $Y = C([a,b],mathbb R)$ into $X = C^1([a,b], mathbb R^n)$ by
                  $T(f)(x) = int_a^x f(t); dt; {bf u}$ where $bf u$ is some nonzero vector in $mathbb R^n$. Then $T^{**}$ embeds $Y^{**}$ into $X^{**}$. But $Y^{**}$ is not separable, therefore $X^{**}$ is not separable, but $X$ is separable, so $X$ is not reflexive.






                  share|cite|improve this answer









                  $endgroup$



                  We can embed $Y = C([a,b],mathbb R)$ into $X = C^1([a,b], mathbb R^n)$ by
                  $T(f)(x) = int_a^x f(t); dt; {bf u}$ where $bf u$ is some nonzero vector in $mathbb R^n$. Then $T^{**}$ embeds $Y^{**}$ into $X^{**}$. But $Y^{**}$ is not separable, therefore $X^{**}$ is not separable, but $X$ is separable, so $X$ is not reflexive.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 27 '18 at 19:16









                  Robert IsraelRobert Israel

                  321k23210462




                  321k23210462






























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